1. More examples of type and cotype
Let be a Banach space. The possible types are in or , . The possible cotypes are in or , .
Cotype does not pass to quotients. Indeed, has cotype , and every separable Banach space is a quotient of . For instance, has trivial cotype.
The type of is .
1.1. Complex interpolation
Type is equivalent to the map , being bounded. By interpolation, if has type and has type , the interpolation space , , has type such that
Cotype does not work so nicely. It is expressible in terms of a map , where
If is complemented, i.e. is the image of a projector independent on (this is the notion of -convexity), interpolation works for cotype.
Clearly, type and cotype are super properties. A stronger property holds.
Definition 1 Say is crudely finitely representable in if there exists such that every finite dimensional subspace of embeds with distorsion in .
The same proof as for ordinary finite representability gives
Proposition 2 is crudely finitely representable in iff is isomorphic to a subspace of an ultrapower .
1.3. Type and finite representability
Lemma 4 Let be independent -valued random variables with a symmetric distribution. Let be independent Bernoulli variables. Then and have the same distribution.
Proof: of Theorem 3. Let denote the best constant in the inequality
Then . Let us show that is submultiplicative, i.e. . Indeed, group variables in packages of .
Assume that for all . So for every , there exist points is close to . Go to an ultrapower to find vectors such that . The equality case in the chain of inequalities leading to shows that all norms and for all choices of signs, . Fix signs . If is a unit functional, , thus . In other words, these functionals give an isometry of span onto .
Otherwise, there exists such that . Then there exists , , such that . Let . Given a family , split it according to norms,
Summing over gives the wanted estimate.
1.4. Ribe’s Theorem
It is our first nonlinear result.
Theorem 5 (Ribe 1976) If is a uniform homeomorphism, then is crudely finitely representable in and is crudely finitely representable in .
Proof: Upgrade the uniform homeomorphism to a bi-Lipschitz map. Use differentiability on finite dimensional subspaces to improve bi-Lipschitz maps to isomorphisms.
This implies that and are not uniformly homeomorphic, an earlier result due to Enflo. Enflo’s proof relies on a metric notion of type, on which Naor and Schechtman, Mendel and Naor have elaborated since.
Definition 6 Let be a metric space. Say has Enflo type if there is a constant such that for all ,
where changes the sign of the -th coordinate.
Remark 1 For a Banach space, Enflo type implies usual type .
1.5. Proof of Ribe’s Theorem
Let be a uniform homeomorphism. Then is large-scale Lipschitz: provided . Go to an ultra product to get rid the the large-scale restriction. This means setting . This is bi-Lischitz: . On every finite dimensional subspace , has points of differentiability. The differential is a linear isomorphism of onto a linear subspace of . So is coarsely finitely representable in , and thus in .
This goes back to Beck and Giesy. They were lookibg for a law of large numbers for Banach space valued random variables.
Lemma 7 Let have type . Let be independent centered -valued random variables. Then
Proof: Reduce to the case of symmetric distributions by replacing by where is an independent copy of . This produces the factor . Then apply Jensen’s inequality
It implies that
which tends to , this is a law of large numbers.
2.1. Application: approximation of convex hulls
Corollary 8 Let be a Banach space with type . Let . Let . There exist such that
Proof: Apply the law of large numbers to independent copies of the following random variable . By assumption, there exist , such that . So let with probability .
The following was one of the first results on isomorphism types.
Theorem 9 (Kwapién) If a Banach space has simultaneously type and cotype , then is is isomorphic to a Hilbert space.
Up to now we have uses Bernoulli averages. Why not use Gaussian averages ? Grouping Bernoulli variables (divided by ) in packages of produces variables which are approximately standard Gaussian (central limit theorem). So we can use Gaussian averages instead.
Lemma 10 Let , be -valued centered Gaussian vectors. Assume that for all ,
Then for every convex function ,
Proof: Pick a common orthogonal basis for the quadratic forms and . Use such a basis to find independant from such that both have the same distribution as .
Let be a finite dimensional normed space. Say a homogeneous function is a cotype function if for all ,
where are independent standard Gaussians.
Lemma 11 There exists a measure on the unit sphere of such that for all ,
Proof: Set ,
These are disjoint convex sets
I have to run to catch the 4:30 train to Orsay…